izfrykse f=kdks.kferh; iqyuncert.nic.in/ncerts/l/lhep202.pdf · iz'u iznf'kzdk &...
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iz'u iznf'kZdk & xf.kr
vè;k; 2
izfrykse f=kdks.kferh; iQyu
2.1 lexz voyksdu (Overview)
2.1.1 izfrykse iQyu
iQyu ‘f ’ osQ izfrykse dk vfLrRo osQoy rHkh gksrk gS tc iQyu ,oSQdh rFkk vkPNknd gks vFkkZr~,oSQdh vkPNknh gks D;ksafd f=kdks.kferh; iQyu cgq,d laxfr (many-one) iQyu gksrs gSa blfy,ge muosQ izkarksa rFkk ifjljksa dks bl izdkj izfrcaf/r djrs gSa fd os ,oSQdh rFkk vkPNknd gks tk,vkSj fiQj ge mudk izfrykse Kkr djrs gSaA izfrykse f=kdks.kferh; iQyuksa osQ izkar rFkk ifjlj (eq[;eku 'kk[kk) uhps fn, x, gSaA
iQyu izkar ifjlj (eq[; eku 'kk[kk)
y = sin–1x [–1,1]–
,2 2
y = cos–1x [–1,1] [0,π]
y = cosec–1x R– (–1,1)–
, – {0}2 2
y = sec–1x R– (–1,1) [0,π] – 2
y = tan–1x R–
,2 2
y = cot–1x R (0,π)
21/04/2018
izfrykse f=kdks.kferh; iQyu 19
fVIi.kh (i) sin–1x ls (sinx)–1 dh Hkzkafr ugha gksuh pkfg,A okLro esa sin–1x ,d dksa.k gS ftlosQ sine
dk eku x gSA ;gh rF; vU; f=kdks.kferh; iQyuksa osQ fy, Hkh lR; gSA
(ii) θ osQ lcls de (U;wure) la[;kRed eku pkgs og /ukRed gks ;k ½.kkRed gks] dksiQyu dk eq[; eku dgrs gSaA
(iii) tc dHkh izfrykse f=kdks.kferh; iQyu dh fdlh fo'ks"k 'kk[kk dk mYys[k u gks rks gekjkrkRi;Z eq[; 'kk[kk ls gksrk gSA izfrykse f=kdks.kferh; iQyu dk og eku tks mldh eq[;'kk[kk osQ ifjlj esa fLFkr gksrk gS mls eq[; eku dgrs gSaA
2.1.2 f=kdks.kferh; iQyuksa dk vkys[kfdlh izfrykse f=kdks.kferh; iQyu dk vkys[k ewy iQyu osQ vkys[k esa x rFkk y-v{kksa dk ijLijfofue; djds izkIr fd;k tk ldrk gSA vFkkZr~] ;fn (a, b) iQyu osQ vkys[k esa ,d fcanq gS rks(b, a) izfrykse iQyu osQ xzkiQ dk laxr fcanq gks tkrk gSA
;g fn[kk;k tk ldrk gS fd izfrykse iQyu osQ vkys[k] js[kk y = x osQ ifjr% laxr ewyiQyu osQ vkys[k dks niZ.k izfrfcac (mirror image) vFkkZr~ ijkorZu (reflection) osQ :i esa izkIrfd;k tk ldrk gSA
2.1.3 izfrykse f=kdks.kferh; iQyuksa osQ xq.k/eZ
1. sin–1 (sin x) = x :–
,2 2
xπ π
∈
cos–1(cos x) = x : [0, ]x π∈
tan–1(tan x) = x :–
,2 2
x
∈
cot–1(cot x) = x : ( )0,x∈
sec–1(sec x) = x :
[0,] –2
x
∈
cosec–1(cosec x) = x :–
, – {0}2 2
x
∈
2. sin (sin–1 x) = x : x ∈[–1,1]
cos (cos–1 x) = x : x ∈[–1,1]
tan (tan–1 x) = x : x ∈R
cot (cot–1 x) = x : x ∈R
21/04/2018
20 iz'u iznf'kZdk
sec (sec–1 x) = x : x ∈R – (–1,1)
cosec (cosec–1 x) = x : x ∈R – (–1,1)
3.–1 –11
sin cosec xx
=
: x ∈R – (–1,1)
–1 –11cos sec x
x
=
: x ∈R – (–1,1)
–1 –11tan cot x
x
=
: x > 0
= – π + cot–1x : x < 0
4. sin–1 (–x) = –sin–1x : x ∈[–1,1]
cos–1 (–x) = π−cos–1x : x ∈[–1,1]
tan–1 (–x) = –tan–1x : x ∈R
cot–1 (–x) = π–cot–1x : x ∈R
sec–1 (–x) = π–sec–1x : x ∈R –(–1,1)
cosec–1 (–x) = –cosec–1x : x ∈R –(–1,1)
5. sin–1x + cos–1x = 2
: x ∈[–1,1]
tan–1x + cot–1x = 2
: x ∈R
sec–1x + cosec–1x = 2
: x ∈R–[–1,1]
6. tan–1x + tan–1y = tan–11 –
x y
xy
+
: xy < 1
tan–1x – tan–1y = tan–1 ; –11
x yxy
xy
−>
+
7. 2tan–1x = sin–12
2
1
x
x+ : –1 ≤ x ≤ 1
2tan–1x = cos–1
2
2
1–
1
x
x+ : x ≥ 0
2tan–1x = tan–12
2
1 –
x
x : –1 < x < 1
21/04/2018
izfrykse f=kdks.kferh; iQyu 21
2.2 gy fd, gq, mnkgj.k
y?kq mÙkjh; (S.A.)
mnkgj.k 1 x = 3
2osQ fy, cos–1x dk eq[; eku Kkr dhft,A
gy ;fn cos–1
3
2
= θ , rc cos θ =
3
2.
ge ;gk¡ eq[; 'kk[kk ij fopkj dj jgs gSa blfy, θ ∈ [0, π]. iqu% 3
2 > 0 ls ge tku x, fd
θ izFke prqFkk±'k eas gS blfy, cos–13
2
= 6
.
mnkgj.k 2 tan–1–
sin2
dks ifjdfyr dhft,A
gy tan–1–
sin2
= tan–1
sin2
−
= tan–1(–1) =
4
− .
mnkgj.k 3 cos–113
cos6
dk eku Kkr dhft,A
gy cos–113
cos6
= cos–1 cos(2 )6
π π+
=
–1 cos cos
6
= 6
π.
mnkgj.k 4 tan–1 9
tan8
dk eku Kkr dhft,A
gy tan–1 9
tan8
= tan–1 tan 8
π π +
= –1tan tan
8
π
=
8
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22 iz'u iznf'kZdk
mnkgj.k 5 tan (tan–1(– 4)) dks ifjdfyr dhft,A
gy D;ksafd x ∈ R osQ lHkh ekuksa osQ fy, tan (tan–1x) = x, gS blfy, tan (tan–1(– 4) = – 4.
mnkgj.k 6 tan–13 – sec–1 (–2) dk eku Kkr dhft,A
gy tan–13 – sec–1 (– 2) = tan–1
3 – [π – sec–12]
= –1 1 2
cos3 2 3 3 3
π π π π −π+ =− + =−
.
mnkgj.k 7 –1 –1 3
sin cos sin2
dk eku Kkr dhft,A
gy –1 –1 –13
sin cos sin sin cos2 3
=
= –1 1
sin2 6
=
.
mnkgj.k 8 fl¼ dhft, fd tan(cot–1x) = cot (tan–1x). dkj.k lfgr crkb, fd D;k ;g x
osQ lHkh ekuksa osQ fy, lR; gSAgy eku yhft, cot–1x = θ. rc cot θ = x
;k,
tan – = 2
x
⇒ –1
tan – 2
x = ;k tan (cot–1x) = –1
tan – tan2
x
= cot (tan–1x)
blfy, –1 –1 –1
tan(cot ) tan cot – cot cot cot(tan )2 2
x x x
= = = − =
;g lerk x osQ lHkh ekuksa osQ fy, lR; gS D;ksafd x ∈ R osQ fy, tan–1x rFkk cot–1x lR; gSA
mnkgj.k 9 sec –1
tan2
y
dk eku Kkr dhft,
gy eku yhft, –1
tan =2
y, tgk¡
,2 2
∈ −
. blfy, , tanθ = 2
y,
ftlls 2
4sec=
2
y+ izkIr gksrk gSA
blfy,] 2
–1 4sec tan =sec=
2 2
yy +
.
21/04/2018
izfrykse f=kdks.kferh; iQyu 23
mnkgj.k 10 tan (cos–1x) dk eku Kkr dhft, vkSj fiQj tan –1 8
cos17
ifjdfyr dhft,A
gy eku yhft, cos–1x = θ, rc cos θ = x, tgk¡ θ ∈ [0,π]
blfy, tan(cos–1x) =
2 21– cos 1–tan =
cosx
x=
vr%
2
–1
81 –
178 15tan cos =
817 8
17
=
mnkgj.k 11 –1 –5
sin 2cot12
dk eku Kkr dhft,
gy eku yhft, cot–1 –5
12
= y . rc cot y = 5
12
−
vc –1 –5
sin 2cot12
= sin 2y
= 2sin y cos y = 12 –5
213 13
cot 0, so ,2
y y
< ∈ D;k fsa d
–120
169=
mnkgj.k 12 –1 –11 4
cos sin sec4 3
+
dk eku Kkr dhft,
gy –1 –11 4
cos sin sec4 3
+
= –1 –11 3
cos sin cos4 4
+
= –1 –1 –1 –11 3 1 3
cos sin cos cos – sin sin sin cos4 4 4 4
=
2 23 1 1 3
1– – 1 –4 4 4 4
21/04/2018
24 iz'u iznf'kZdk
= 3 15 1 7 3 15 – 7
–4 4 4 4 16
=
nh?kZ mÙkjh; mÙkj (L.A.)
mnkgj.k 13 fl¼ dhft, fd 2sin–13
5 – tan–1
17
31 =
4
π
gy eku yhft, sin–1 3
5= θ, rc sinθ =
3
5, tgk¡ θ ∈ ,
2 2
−π π
bl izdkj tan θ = 3
4, ftlls θ = tan–1
3
4 izkIr gksrk gSA
blfy, 2sin–13
5 – tan–1
17
31
= 2θ – tan–1 17
31 = 2 tan–1
3
4 – tan–1
17
31
= –1 –1
32.
174tan – tan9 31
1–16
= tan–1 –124 17
tan7 31
−
= –1
24 17
7 31tan24 17
1 .7 31
−
+
= 4
π
mnkgj.k 14 fl¼ dhft, fdcot–17 + cot–18 + cot–118 = cot–13
gy fn;k gS cot–17 + cot–18 + cot–118
= tan–11
7 + tan–1
1
8 + tan–1
1
18 (D;ksafd x > 0 osQ fy, cot–1 x = tan–1
1
x)
21/04/2018
izfrykse f=kdks.kferh; iQyu 25
= –1 –1
1 1
17 8tan tan1 1 18
17 8
+
+ − ×
(D;ksafd x . y = 1 1
.7 8
< 1)
= –1 –13 1
tan tan11 18
+ = –1
3 1
11 18tan3 1
111 18
+
− ×
(D;ksafd xy < 1)
= –1 65
tan195
= –1 1
tan3
= cot–1 3
mnkgj.k 15 tan 1 rFkk tan–1 1 esa ls dkSu lk cM+k gS?
gy vko`Qfr 2-1 ls ge ns[krs gSa
fd varjky ,2 2
−π π
es a tan x
o/Zeku iQyu gSA D;ksafd
1 > 4
π ⇒ tan 1 > tan
4
π
vr%] tan 1 > 1
⇒ tan 1 > 1 > 4
π
⇒ tan 1 > 1 > tan–1 (1).
mnkgj.k 16 –1 –12
sin 2tan cos(tan 3)3
+
dk eku Kkr dhft,
gy ekuk tan–1 2
3 = x vkSj tan–1 3 = y blfy, tan x =
2
3 vkSj tan y = 3
vr%] –1 –12
sin 2tan cos(tan 3)3
+
= sin (2x) + cos y
vko`Qfr 2-1
21/04/2018
26 iz'u iznf'kZdk
= 2 2
2 tan 1
1 tan 1 tan
x
x y+
+ + = ( )
2
22.
134
1 319
+++
= 12 1 37
13 2 26+ = .
mnkgj.k 17 –1 –11 1
tan tan , 01 2
xx x
x
−= >
+ dks x osQ fy, gy dhft,
gy fn, x, lehdj.k ls] –1 –112tan tan
1
xx
x
−=
+
–1 –1 –12 tan 1 tan tanx x − =
–1
2 3tan4
xπ
=
–1tan6
xπ
=
1
3x = .
mnkgj.k 18 x osQ os eku Kkr dhft, tks lehdj.k sin–1 x + sin–1 (1 – x) = cos–1 x dks larq"V
djrs gSaA
gy fn, x, lehdj.k ls gesa izkIr gksrk gS fd
sin [sin–1 x + sin–1 (1 – x)] = sin (cos–1x)
sin (sin–1 x) cos (sin–1 (1 – x)) + cos (sin–1 x) sin (sin–1 (1 – x) ) = sin (cos–1 x)
2 2 21– (1– ) (1 ) 1 1x x x x x+ − − = −
2 22 – 1 (1 1) 0x x x x x+ − − − =
( )2 22 – 1 0x x x x− − =
x = 0 ;k 2x – x2 = 1 – x2
x = 0 ;k x = 1
2.
21/04/2018
izfrykse f=kdks.kferh; iQyu 27
mnkgj.k 19 lehdj.k sin–16x + sin–1 6 3 x = 2
π− dks gy dhft,A
gy fn, x, lehdj.k dks sin–1 6x = –1
sin 6 32
xπ
− − osQ :i esa fy[k ldrs gSaA
⇒ sin (sin–1 6x) = sin –1
sin 6 32
xπ
− −
⇒ 6x = – cos (sin–1 6 3 x)
⇒ 6x = – 21 108x−
oxZ djus ij izkIr gksrk gS 36x2 = 1 – 108x2
⇒ 144x2 = 1 ⇒ x = ± 1
12
è;ku nhft, fd osQoy x = – 1
12 gh lehdj.k dk gy gS D;ksafd x =
1
12 bls larq"V ugha djrk gSA
mnkgj.k 20 n'kkZb, fd
2 tan–1 –1 sin cos
tan .tan tan2 4 2 cos sin
α π β α β − =
α + β
gy L.H.S. = –1
2 2
2 tan .tan2 4 2
tan
1 tan tan2 4 2
α π β −
α π β − −
–1 –1
2
22 tan tan
1
xx
x
=
− D;kafs d
= –1
2
2
1 tan22tan
21 tan
2tan
1 tan21 tan
21 tan
2
β − α β +
β − α
− β +
21/04/2018
28 iz'u iznf'kZdk
=
2
–1
2 2
2
2 tan . 1 tan2 2
tan
1 tan tan 1 tan2 2 2
α β −
β α β
+ − −
=
2
–1
2 2 2
2 tan 1 tan2 2
tan
1 tan 1 tan 2 tan 1 tan2 2 2 2
α β −
β α β α
+ − + +
=
2
2 2
–1
2
2 2
2 tan 1 tan2 2
1 tan 1 tan2 2tan
1 tan 2 tan2 2
1 tan 1 tan2 2
α β−
α β+ +
α β−
+α β
+ +
=–1 sin cos
tancos sin
α β
α+ β = R.H.S.
cgqfodYih; iz'u (M.C.Q.)
iz'u 21 ls 41 rd izR;sd osQ fy, fn, x, pkj fodYiksa esa ls lgh fodYi pqfu,&
mnkgj.k 21 fuEu esa ls dkSu lk tan–1 dh eq[; eku 'kk[kk gS?
(A) ,2 2
π π −
(B) ,2 2
π π −
(C) ,2 2
π π −
– {0} (D) (0, π)
gy lgh mÙkj (A) gSA
mnkgj.k 22 sec–1 dh eq[; eku 'kk[kk gSA
(A) { }, 02 2
π π − −
(B) [ ]0,2
π π −
(C)(0, π) (D) ,
2 2
π π −
gy lgh mÙkj (B) gSA
21/04/2018
izfrykse f=kdks.kferh; iQyu 29
mnkgj.k 23 eq[; eku 'kk[kk osQ vfrfjDr cos–1 dh ,d vU; 'kk[kk gS
(A)3
,2 2
π π
(B) [ ]3
, 22
π π π −
(C) (0, π) (D) [2π, 3π]
gy lgh mÙkj (D) gSA
mnkgj.k 24 –1 43
sin cos5
π
dk eku gS
(A)3
5
π(B)
7
5
− π(C)
10
π(D) –
10
π
gy lgh mÙkj (D) gSA D;ksafd –1 –140 3 3sin cos sin cos 8
5 5
π+ π π = π+
= –1 –13 3
sin cos sin sin5 2 5
π π π = −
= –1sin sin
10 10
π π − = −
.
mnkgj.k 25 O;atd cos–1 [cos (– 680°)] dk eku gS
(A)2
9
π(B)
2
9
− π(C)
34
9
π(D)
9
π
gy lgh mÙkj (A) gS D;ksafd cos–1 (cos (680°)) = cos–1 [cos (720° – 40°)]
= cos–1 [cos (40°)] = 40° = 2
9
π.
mnkgj.k 26 cot (sin–1x) dk eku gS
(A)2
1 x
x
+(B) 2
1
x
x+(C)
1
x (D)
21 x
x
−
gy lgh mÙkj (D) gSA eku yhft, sin–1 x = θ, rc sinθ = x
21/04/2018
30 iz'u iznf'kZdk
⇒ cosec θ = 1
x⇒ cosec2θ = 2
1
x
⇒ 1 + cot2 θ = 2
1
x⇒ cotθ =
21 x
x
−.
mnkgj.k 27 ;fn fdlh x ∈ R osQ fy, tan–1x = 10
π gS rks cot–1x dk eku gS
(A)5
π(B)
2
5
π(C)
3
5
π(D)
4
5
π
gy lgh mÙkj (B) gSA ge tkurs gSa fd tan–1x + cot–1x = 2
π blfy, cot–1x =
2
π –
10
π
⇒ cot–1x = 2
π –
10
π =
2
5
π.
mnkgj.k 28 sin–1 2x dk izkar gS
(A) [0, 1] (B) [– 1, 1] (C)1 1
,2 2
−
(D) [–2, 2]
gy lgh mÙkj (B) gSA eku yhft, sin–12x = θ ;k 2x = sin θ.
vc – 1 ≤ sin θ ≤ 1, vFkkZr~ – 1 ≤ 2x ≤ 1 ftlls 1 1
2 2x− ≤ ≤ izkIr gksrk gSA
mnkgj.k 29 sin–1 3
2
−
dk eq[; eku gS
(A)2
3
π− (B)
3
π− (C)
4
3
π(D)
5
3
π
gy lgh mÙkj (B) gSA D;ksafd
–1 –1 –13sin sin –sin –sin sin –
2 3 3 3
− π π π = = =
.
mnkgj.k 30 (sin–1x)2 + (cos–1x)2 dk Øe'k% vf/dre rFkk U;wure eku gS
21/04/2018
izfrykse f=kdks.kferh; iQyu 31
(A)
2 25
4 8
π πrFkk (B)
2 2
π −πrFkk (C)
2 2
4 4
π −πrFkk (D)
2
04
πrFkk
gy lgh mÙkj (A) gSA ge tkurs gSa fd
(sin–1x)2 + (cos–1x)2 = (sin–1x + cos–1x)2 – 2 sin–1x cos–1 x
=
2–1 –12sin sin
4 2x x
π π − −
= ( )2
2–1 –1
sin 2 sin4
x xπ
− π +
= ( )2
2–1 –1
2 sin sin2 8
x x π π
− +
=
2 2–12 sin
4 16x
π π − +
.
bl izdkj] U;wure eku 2 2
216 8
π π
vFkkZr ~ gS rFkk vf/dre eku
2 2
22 4 16
−π π π − +
,
vFkkZr~
25
4
π gSA
mnkgj.k 31 ;fn θ = sin–1 (sin (– 600°), rc θ dk eku gS
(A)3
π(B)
2
π(C)
2
3
π(D)
2
3
− π
gy lgh mÙkj (A) gS D;ksafd
–1 –1 10
sin sin 600 sin sin180 3
π − π − × =
= –1 2
sin sin 43
π − π−
= –1 2
sin sin3
π
21/04/2018
32 iz'u iznf'kZdk
= –1 –1
sin sin sin sin3 3 3
π π π π− = =
.
mnkgj.k 32 iQyu y = sin–1 (– x2) dk izkar gS(A) [0, 1] (B) (0, 1) (C) [–1, 1] (D) φ
gy lgh mÙkj (C) gS D;ksafd y = sin–1 (– x2)⇒ siny = – x2
vFkkZr~ – 1 ≤ – x2 ≤ 1 (D;ksafd – 1 ≤ sin y ≤ 1)
⇒ 1 ≥ x2 ≥ – 1
⇒ 0 ≤ x2 ≤ 1
⇒ 1 1 1x x≤ − ≤ ≤;k
mnkgj.k 33 y = cos–1 (x2 – 4) dk izkar gS(A) [3, 5] (B) [0, π]
(C) 5, 3 5, 3 − − ∩ − (D) 5, 3 3, 5 − − ∪
gy lgh mÙkj (D) gS D;ksafd y = cos–1 (x2 – 4 )⇒ cosy = x2 – 4
vFkkZr~ – 1 ≤ x2 – 4 ≤ 1 (D;ksafd – 1 ≤ cos y ≤ 1)
⇒ 3 ≤ x2 ≤ 5
⇒ 3 5x≤ ≤
⇒ 5, 3 3, 5x ∈ − − ∪
mnkgj.k 34 f (x) = sin–1x + cosx }kjk ifjHkkf"kr iQyu dk izkar gS
(A) [–1, 1] (B) [–1, π + 1] (C) ( )– ,∞ ∞ (D) φ
gy lgh mÙkj (A) gSA D;ksafd iQyu cos dk izkar R gS rFkk sin–1 dk izkar [–1, 1] gSA blfy,
f (x) = cosx + sin–1x dk izkar R [ ]–1,1∩ , vFkkZr~ [–1, 1] gSA
mnkgj.k 35 sin (2 sin–1 (.6)) dk eku gS
(A) .48 (B) .96 (C) 1.2 (D) sin 1.2
gy lgh mÙkj (A) gSA ;fn sin–1 (0.6) = θ, rc sin θ = .6.
vc sin (2θ) = 2 sinθ cosθ = 2 (.6) (.8) = .96
21/04/2018
izfrykse f=kdks.kferh; iQyu 33
mnkgj.k 36 ;fn sin–1 x + sin–1 y = 2
π, rc cos–1 x + cos–1 y dk eku gS
(A)2
π(B) π (C) 0 (D)
2
3
π
gy lgh mÙkj (B) gSA D;ksafd sin–1 x + sin–1 y = 2
π gS blfy,
–1 –1– cos – cos
2 2 2x y
π π π + =
⇒ cos–1x + cos–1y = 2
π.
mnkgj.k 37 tan –1 –13 1
cos tan5 4
+
dk eku gS
(A)19
8(B)
8
19(C)
19
12(D)
3
4
gy lgh mÙkj (A) gSA D;ksafd tan –1 –13 1
cos tan5 4
+
= tan
–1 –14 1tan tan
3 4
+
= tan tan –1
–1
4 119 193 4 tan tan
4 1 8 81
3 4
+
= = − ×
.
mnkgj.k 38 O;atd sin [cot–1 (cos (tan–1 1))] dk eku gS
(A) 0 (B) 1 (C)1
3(D)
2
3
gy lgh mÙkj (D) gSA D;ksafd
sin [cot–1 (cos 4
π)] = sin [cot–1
1
2]=
–1 2 2sin sin
3 3
=
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34 iz'u iznf'kZdk
mnkgj.k 39 lehdj.k tan–1x – cot–1x = tan–1 1
3
(A) dk dkbZ gy ugha gS (B) dk osQoy ,d ek=k gy gS(C) osQ vuar gy gSa (D) osQ nks gy gSa
gy lgh mÙkj (B) gSA D;ksafd
tan–1x – cot–1x = 6
π rFkk tan–1x + cot–1x =
2
π.
budks tksM+us ij gesa 2tan–1x = 2
3
π izkIr gksrk gS
blfy, ⇒ tan–1x = 3
π vFkkZr~ 3x= .
mnkgj.k 40 ;fn 2α≤ sin–1x + cos–1x ≤β , rc
(A) ,2 2
−π πα = β= (B) 0,α = β=π
(C)3
,2 2
−π πα= β= (D) 0, 2α = β= π
gy lgh mÙkj (B) gSA fn;k x;k gS fd 2
−π≤ sin–1 x ≤
2
π
⇒2
−π+
2
π ≤ sin–1x +
2
π ≤
2
π +
2
π
⇒ 0 ≤ sin–1x + (sin–1x + cos–1x) ≤ π
⇒ 0 ≤ 2sin–1x + cos–1x ≤ π
mnkgj.k 41 tan2 (sec–12) + cot2 (cosec–13) dk eku gS(A) 5 (B) 11 (C) 13 (D) 15
gy lgh mÙkj (B) gSA
tan2 (sec–12) + cot2 (cosec–13) = sec2 (sec–12) – 1 + cosec2 (cosec–13) – 1
= 22 × 1 + 32 – 2 = 11.
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izfrykse f=kdks.kferh; iQyu 35
2.3 iz'ukoyh
y?kq mÙkjh; iz'u (S.A.)
1.–1 –15 13
tan tan cos cos6 6
+
dk eku fudkfy,A
2.–1 – 3
cos cos2 6
π
+
dk eku Kkr dhft,A
3. fl¼ dhft, fd –1cot – 2 cot 3 7
4
π =
.
4.–1 –1 –11 1 –
tan – cot tan sin23 3
π + +
dk eku fudkfy,A
5. tan–1 2
tan3
dk eku fudkfy,A
6. n'kkZb, fd 2tan–1 (–3) = –
2
π +
–1 –4tan
3
.
7. lehdj.k ( )–1 –1 2 tan 1 sin 1
2x x x x+ + + + = . osQ okLrfod gy Kkr dhft,A
8. O;atd sin ( )–1 –112 tan cos tan 2 2
3
+
dk eku fudkfy,A
9. ;fn 2 tan–1 (cos θ) = tan–1 (2 cosec θ), rks fn[kkb, fd θ = 4
.
10. n'kkZb, fd–1 –11 1
cos 2tan sin 4 tan7 3
=
.
11. lehdj.k ( )–1 –1 3cos tan sin cot
4x
=
dks gy dhft,
21/04/2018
36 iz'u iznf'kZdk
nh?kZ mÙkjh; iz'u (L.A.)
12. fl¼ dhft, fd
2 2
–1 –1 2
2 2
1 1– 1tan cos
4 21 – 1–
x xx
x x
π + += +
+
.
13.–1 3 4
cos cos sin5 5
x x
+
, tgk¡ x ∈ –3
,4 4
π π
, dks ljyre :i esa fyf[k,A
14. fl¼ dhft, fd –1 –1 –18 3 77sin sin sin
17 5 85+ = .
15. n'kkZb, fd –1 –1 –15 3 63
sin cos tan13 5 16
+ = .
16. fl¼ dhft, fd –1 –1 11 2 1
tan tan sin4 9 5
−+ = .
17.–1 –11 1
4 tan – tan5 239
dk eku Kkr dhft,A
18. n'kkZb, fd –11 3 4 – 7tan sin
2 4 3
=
rFkk bldk Hkh vkSfpR; crkb, fd nwljk eku
4 7
3
+ dks D;ksa ugha fy;k x;k gSA
19. ;fn a1, a
2, a
3,...,a
n ,d lekarj Js<+h esa gSa ftldk lkoZ varj (common difference)
d gS rks fuEufyf[kr O;atd dk eku fudkfy,A
–1 –1 –1 –1
1 2 2 3 3 4 –1
tan tan tan tan ... tan1 1 1 1 n n
d d d d
a a a a a a a a
+ + + +
+ + + + .
cgqfodYih; iz'u (M.C.Q.)
iz'u 20 ls 37 rd izR;sd osQ fy, fn, x, pkj fodYiksa esa ls lgh fodYi pqfu,&20. fuEu esa ls dkSu lk cos–1x dh eq[; 'kk[kk gS?
(A) –
,2 2
(B) (0, π) (C) [0, π] (D) (0, π) – 2
21/04/2018
izfrykse f=kdks.kferh; iQyu 37
21. fuEufyf[kr esa ls dkSu lk cosec–1x dh eq[; 'kk[kk gS?
(A) –
,2 2
(B) [0, π] – 2
(C) –
,2 2
(D)–
,2 2
– {0}
22. ;fn 3tan–1 x + cot–1 x = π, rks x cjkcj gS
(A) 0 (B) 1 (C) –1 (D) 1
2
23. sin–1 33
cos5
π
dk eku gS
(A) 35
(B) –7
5
π(C)
10
π(D)
–
10
π
24. iQyu cos–1 (2x – 1) dk izkar gS(A) [0, 1] (B) [–1, 1] (C) ( –1, 1) (D) [0, π]
25. f (x) = sin–1 –1x }kjk ifjHkkf"kr iQyu dk izkar gS
(A) [1, 2] (B) [–1, 1] (C) [0, 1] (D) buesa ls dksbZ ugha
26. ;fn cos –1 –12
sin cos 05
x
+ =
, rks x dk eku gS
(A) 1
5(B)
2
5(C) 0 (D) 1
27. sin (2 tan–1 (.75)) dk eku gS(A) .75 (B) 1.5 (C) .96 (D) sin 1.5
28.–1 3
cos cos2
π
dk eku gS
(A) 2
π(B)
3
2
π(C)
5
2
π(D)
7
2
π
29. O;atd 2 sec–1 2 + sin–1 1
2
dk eku gS
(A) 6
(B) 56
(C) 76
(D) 1
21/04/2018
38 iz'u iznf'kZdk
30. ;fn tan–1 x + tan–1y = 45
, rks cot–1 x + cot–1 y cjkcj gS
(A) 5
(B) 25
(C) 3
5
π(D) π
31. ;fn sin–1
2–1 –1
2 2 2
2 1– 2cos tan
1 1 1–
a a x
a a x
+ =
+ + , tgk¡ a, x ∈ ]0, 1, rc x
dk eku cjkcj gS
(A) 0 (B) 2
a(C) a (D) 2
2
1–
a
a
32. cot –1 7
cos25
dk eku gS
(A) 25
24(B)
25
7(C)
24
25(D)
7
24
33. O;atd tan –11 2
cos2 5
dk eku gS
(A) 2 5+ (B) 5 – 2 (C) 5 2
2
+(D) 5 2+
1– cos: tan
2 1 cos
θ θ=
+ θ laoQs r iz;qDr djsa
34. ;fn | x | ≤ 1, rc 2 tan–1 x + sin–12
2
1
x
x
+
cjkcj gS
(A) 4 tan–1 x (B) 0 (C) 2
π(D) π
35. ;fn cos–1 α + cos–1 β + cos–1 γ = 3π, rc α (β + γ) + β (γ + α) + γ (α + β) cjkcj gS(A) 0 (B) 1 (C) 6 (D) 12
21/04/2018
izfrykse f=kdks.kferh; iQyu 39
36. lehdj.k –1
1 cos 2 2 cos (cos )in ,2
x xπ
+ = π osQ okLrfod gyksa dh la[;k gS
(A) 0 (B) 1 (C) 2 (D) vuar
37. ;fn cos–1x > sin–1x, gks rks
(A) 1
12
x< ≤ (B) 1
02
x≤ < (C) 1
12
x− ≤ < (D) x > 0
iz'u 38 ls 48 rd fjDr LFkku Hkfj, &
38. cos–1 1
–2
dh eq[; 'kk[kk __________ gSA
39. sin–1 3
sin5
π
dk eku __________gSA
40. ;fn cos (tan–1 x + cot–1 3 ) = 0, rc x dk eku __________ gSA
41. sec–1 1
2
osQ ekuksa dk leqPp; __________ gSA
42. tan–1 3 dk eq[; eku __________gSA
43. cos–1 14
cos3
π
dk eku __________ gSA
44. cos (sin–1 x + cos–1 x), |x| ≤ 1 dk eku __________ gSA
45. O;atd tan
–1 –1sin cos
2
x x +
,tgk¡ x = 3
2 gS] dk eku __________ gSA
46. ;fn x osQ lHkh ekuksa osQ fy, y = 2 tan–1 x + sin–12
2
1
x
x
+
rc ____< y <____.
47. ifj.kke tan–1x – tan–1y = tan–1 1
x y
xy
−
+ rHkh lR; gS tc xy _____ gSA
48. lHkh x ∈ R osQ fy, cot–1 (–x) dk eku cot–1x osQ in esa _______ gSA
21/04/2018
40 iz'u iznf'kZdk
iz'u 49 ls 55 rd izR;sd esa fn, x, dFku dks crkb, fd og lR; gS ;k vlR;&
49. izR;sd f=kdks.kferh; iQyu dk muosQ laxr iazkrksa esa izfrykse iQyu dk vfLrRo gksrk gSA
50. O;atd (cos–1 x)2 dk eku sec2 x osQ cjkcj gSA
51. f=kdks.kferh; iQyuksa osQ iazkrksa dk mudh fdlh Hkh 'kk[kk (vko';d ugha fd eq[; 'kk[kkgks) esa izfrcaf/r fd;k tk ldrk gS rkfd mudk izfrykse iQyu izkIr gks losQA
52. θ dks.k dk U;wure la[;kRed eku] pkgs /ukRed gks ;k ½.kkRed] dks f=kdks.kferh;iQyu dk eq[; eku dgrs gSaA
53. izfrykse f=kdks.kferh; iQyuksa dk vkys[k muosQ laxr f=kdks.kferh; iQyu osQ vkys[k esax rFkk y v{k dk ijLij fofue; djosQ izkIr fd;k tk ldrk gSA
54. n dk og U;wure eku ftlosQ fy, tan–1 ,4
nn
π> ∈
πN , osQ fy, lR; gks] og 5 gSA
55. sin–1 –1 1
cos sin2
dk eq[; eku 3
π gSA
21/04/2018